Levi-civita permutation tensor, and kroneker delta

Hello, I'm interested in seeing some proof of the identities involving the levi civita permutation tensor and and the kroneker delta. I've discovered the utility and efficiency of these identities in deriving the standard vector calculus identities involving div, grad, and curl, but I'm sort of just applying a formula which I am taking on faith in the process.

I have no formal knowledge of tensors, tensor calculus and the like, and little formal linear algebra knowledge.

Is it possible to find a proof of these identities which doesn't involve one or the other, or one which is semi-convincing that I can satisfy myself with before taking on the subjects of linear algebra and tensor analysis?

Thanks!

Hello, can anyone help me?
I have to solve this 3-dimension integral:
ui*uj*uk*ul du
where u is a versor.
Is it equal to:
delta(i,j)*delta(k,l)+delta(i,k)*delta(j,l)+delta(i,l)*delta(j,k)?
where delta=delta Kronecker

if yes what about the the integrals:

ui*uj*uk*ul*um du
and
ui*uj*uk*ul*um*un du?

(ui,uj,ul,um,un generic component of the versor u)

. I apologize if I don't use Latex (it isn't my pc)

Thanks

silvia

dextercioby
Homework Helper
Well the versor must be written in a basis and therefore its components in that basis must be written in any integral.

IF you know the LaTex code you could just type formulas inside [ tex ] tags.

Daniel.

Hi, the integral that I have to solve is this:
$$\begin{equation} \int{d^3u u_a u_b u_c u_d u_e u_f} \end{equation}[tex] If I have the T^6 of the Legendre polynom in three dimension all would be done! silvia sorry I wrong [tex]!! Hi, the integral that I have to solve is this: [tex] \int{d^3u u_a u_b u_c u_d u_e u_f}$$
If I have the T^6 of the Legendre polynom in three dimension all would be done!

silvia

I wronged again...now maybe!
Hi, the integral that I have to solve is this:
$$\begin{equation} \int{d^3u u_a u_b u_c u_d u_e u_f} \end{equation}$$
If I have the T^6 of the Legendre polynom in three dimension all would be done!

silvia